Swap Two Variables - Problem

Write a program to swap two variables without using a third variable.

You are given two integers and need to exchange their values without creating any additional temporary variable. There are multiple mathematical and bitwise approaches to achieve this.

Your function should take two integers as input and return them in swapped order as an array [b, a] where the original values were a and b.

Input & Output

Example 1 — Basic Swap

$ Input: a = 5, b = 3

› Output: [3, 5]

💡 Note: The values are swapped: original a=5 becomes second element, original b=3 becomes first element

Example 2 — Negative Numbers

$ Input: a = -2, b = 7

› Output: [7, -2]

💡 Note: Works with negative numbers: original a=-2 becomes second, original b=7 becomes first

Example 3 — Same Values

$ Input: a = 4, b = 4

› Output: [4, 4]

💡 Note: When both values are same, swapping still works and result is [4, 4]

Constraints

-10⁹ ≤ a, b ≤ 10⁹
Must not use any temporary variables
Return format: [b, a] where original inputs were a and b

Visualization

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Asked in

M Microsoft 25 a Amazon 20 G Google 15

The key insight is to use mathematical properties like addition/subtraction or XOR to temporarily encode both values in one variable. Best approach is XOR manipulation as it avoids overflow risks. Time: O(1), Space: O(1)

Common Approaches

✓ Addition and Subtraction

⏱️ Time: O(1) Space: O(1)

Store the sum of both numbers in the first variable, then use subtraction to extract the original values. This approach uses mathematical properties to avoid needing a third variable.

XOR Bit Manipulation

⏱️ Time: O(1) Space: O(1)

Leverage the mathematical property that X ⊕ X = 0 and X ⊕ 0 = X. By applying XOR operations in sequence, we can extract the original values without needing extra storage.

Addition and Subtraction — Algorithm Steps

Store sum of both numbers in first variable
Extract second number by subtracting from sum
Extract first number by subtracting from sum

Visualization

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Step-by-Step Walkthrough

Store Sum

a = a + b (keeps total)

Extract B

b = a - b (gets original a)

Extract A

a = a - b (gets original b)

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void solution(int* a, int* b) {
    *a = *a + *b;  // a now contains sum
    *b = *a - *b;  // b now contains original a
    *a = *a - *b;  // a now contains original b
}

int main() {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    solution(&a, &b);
    printf("[%d,%d]\n", a, b);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Only three arithmetic operations regardless of input size

✓ Linear Growth

Space Complexity

O(1)

No additional variables or data structures used

✓ Linear Space

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Swap Two Variables - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Addition and Subtraction — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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